Octane (C3H18) undergoes combustion according to the following thermochemical equation:2C3H18(1) + 2502(g)16CO2(g) + 18H20(1)°pxn = -11,020 kJ/mol.Given that AH°{CO2(g)] = -393.5 kJ/mol and AH°{{H2O(1)] = -285.8 kJ/mol, calculate the standard enthalpy of formation of octane.O -420 kJ/molO 22,040 kJ/molO - 210 kJ/molO 420 kJ/molO -11,230 kJ/mol

Given chemical reaction : 2C8H18 + 25 O2 ---> 16 CO2 + 18 H2O

C3H18(1) + 2502(g) - 16CO2(9) + 18H20(1) AH°n = -11,020 KJ/mol. iven that AH°{CO2(9)] = -393.5 kJ/mol and AH?{H2O(1)] = -285.8 kJ/mol, calculate the standard enthalpy of formatio O -420 kJ/mol O 22,040 kJ/mol O -210 kJ/mol O 420 kJ/mol O -11,230 kJ/mol

C3H18(1) + 2502(g) - 16CO2(9) + 18H20(1) AH°n = -11,020 KJ/mol. iven that AH°{CO2(9)] = -393.5 kJ/mol and AH?{H2O(1)] = -285.8 kJ/mol, calculate the standard enthalpy of formatio O -420 kJ/mol O 22,040 kJ/mol O -210 kJ/mol O 420 kJ/mol O -11,230 kJ/mol

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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